# Show that $\frac{P(z)}{Q(z)} = \sum_{k=1}^{n}\frac{P(\alpha_k)}{Q'(\alpha_k)(z-\alpha_k)}$

Here, $Q$ is a polynomial with distinct roots $\alpha_1, \ldots, \alpha_n$ and $P$ is a polynomial of degree $<n$. Once again, the task is to show

$$\frac{P(z)}{Q(z)} = \sum_{k=1}^{n}\frac{P(\alpha_k)}{Q'(\alpha_k)(z-\alpha_k)}$$

For reference, this is page 32 exercise 2 in Complex Analysis by Ahlfors. I'm having a ton of trouble making connections between what is presented in the text and this problem. The section is on rational functions and is incredibly concise. There was never a mention of derivatives and their relationships to rational functions, so I'm guessing here that it just means that $Q'(a_k)$ is a polynomial of degree $n-1$.

I don't even know where to start here, but I do have a couple question which might motivate the proof. What does $P(\alpha_k)$ mean? I don't see what a root of $Q$ has to do with a completely different polynomial. What is the relationship between the roots of $Q$ and the roots of $Q'$? And finally, where does the derivative of $Q$ come in?

• Hint: move $Q(z)$ to the other side, then prove that the equality holds for each $z=\alpha_k\,$. Though, in fairness, that won't come easy if you really know nothing about derivatives and what $Q'$ means.
– dxiv
Feb 27, 2017 at 4:43
• Well, I know as much about derivatives of functions with complex coefficients as is presented (that I can understand without the aid of a classroom) in the analytic functions and polynomials sections in Ahlfors, which are the two sections preceding this one. There is a sentence which says that $P(\alpha) = P'(\alpha) = \cdots = P^{(h-1)}(\alpha) = 0$, so I suppose then that $Q'(a_k) = 0$ given that $n>k$? I am still unsure how to relate $\alpha_k$ to $P$. Thank you for the hint. Feb 27, 2017 at 5:05
• Elaborated some more in the posted answer. Q′(a_k)=0 given that n>k? No, $Q′(\alpha_k)$ is the value of the derivative $Q'(z)$ at $z=\alpha_k$. still unsure how to relate α_k to P They are not directly "related" except by the given identity. $P'(\alpha_k)$ is just the value $P(z)$ takes when $z=\alpha_k\,$.
– dxiv
Feb 27, 2017 at 5:13

Hint (without L'Hôpital):  assume WLOG $Q$ is monic (otherwise cancel out the leading term between the two sides). Then $Q(z)=\prod_{k=1}^n(z-\alpha_k)$ and, by the product rule of differentiation, $\,Q'(z)=\sum_{k=1}^n \prod_{j \ne k}(z-\alpha_j)\,$ so in particular $\,Q'(\alpha_k)=\prod_{j \ne k}(\alpha_k-\alpha_j)\,$.

Multiplying by $Q(z)$ and using the above, the equality to prove becomes:

\begin{align} P(z) = \sum_{k=1}^{n}\frac{P(\alpha_k)\,Q(z)}{Q'(\alpha_k)(z-\alpha_k)} & = \sum_{k=1}^{n}\frac{P(\alpha_k)\,Q(z)}{Q'(\alpha_k)(z-\alpha_k)} \\[5px] & = \sum_{k=1}^{n} \frac{P(\alpha_k)\,\prod_{j \ne k}(z-\alpha_j)}{\prod_{j \ne k}(\alpha_k-\alpha_j)} \\[5px] \end{align}

It follows that the equality holds for all $\,z=\alpha_k\,$ and, since the two sides of the equality are polynomials of degree $\le n-1\,$ that are equal at $n$ points, it further follows that the equality holds in general, for all $z$.

• This is the Lagrange Polynomial matching at the $a_k$.
– robjohn
Feb 27, 2017 at 5:11
• @robjohn It is indeed, thanks for pointing out. The above is (quite obviously) a more painful derivation than yours. Yet, given the little context provided, I thought a purely algebraic alternative might be worth mentioning.
– dxiv
Feb 27, 2017 at 5:17
• Thank you for this! I was, surprisingly, able to follow everything, but I still must think about why the conclusion follows. I apologize for my lack of knowledge (and hence the little context) on the topic. My only source is this textbook, and it's challenging for me. I have one last question: following up from your comment: if $a_k$ is a root for $Q$, why does $Q'(a_k) \neq 0$? Feb 27, 2017 at 5:19
• @playitright $Q'(\alpha_k) \ne 0$ because all roots are assumed to be simple. If you ask why it's allowed to multiply/divide by $Q(z)$ despite it being zero at the $n$ points $\alpha_k\,$, then the answer is a bit more complicated. If taken as abstract "fractions" of polynomials, then there is no problem. Else if taken as complex functions, note that both sides of the original equality are technically undefined for $z=\alpha_k\,$. The identity can be extended to the entire $\mathbb{C}$ by continuity, though, and the same argument could be used with the proof above.
– dxiv
Feb 27, 2017 at 5:35
• @playitright: If $Q(x)=Q'(x)=0$, then $x$ is a double root of $Q$. The question stated that $Q$ has distinct roots.
– robjohn
Feb 27, 2017 at 5:35

Use Partial Fractions and L'Hôpital: $$\frac{P(x)}{Q(x)}=\sum_{k=1}^n\frac{A_k}{x-a_k}\\$$ where $A_k=\lim\limits_{x\to a_k}(x-a_k)\frac{P(x)}{Q(x)}=\frac{P(a_k)}{Q'(a_k)}$.

• Curious that the zeroes of $P$ are implicitly embedded in the expansion. It might be worthwhile mentioning this. Feb 27, 2017 at 4:55